Solve this

Solution:

Let, $(a+i b)^{2}=16-30 i$

Now using, $(a+b)^{2}=a^{2}+b^{2}+2 a b$

$\Rightarrow a^{2}+(b i)^{2}+2 a b i=16-30 i$

Since $i^{2}=-1$

$\Rightarrow a^{2}-b^{2}+2 a b i=16-30 i$

Now, separating real and complex parts, we get

$\Rightarrow a^{2}-b^{2}=16$ …………..eq.1

$\Rightarrow 2 a b=-30$ …….. eq.2

$\Rightarrow \mathrm{a}=-\frac{15}{b}$

Now, using the value of a in eq.1, we get

$\Rightarrow\left(-\frac{15}{b}\right)^{2}-b^{2}=16$

$\Rightarrow 225-b^{4}=16 b^{2}$

$\Rightarrow b^{4}+16 b^{2}-225=0$

Simplify and get the value of $b^{2}$, we get,

$\Rightarrow b^{2}=-25$ or $b^{2}=9$

As $b$ is real no. so, $b^{2}=9$

$\mathrm{b}=3$ or $\mathrm{b}=-3$

Therefore, $a=-5$ or $a=5$

Hence the square root of the complex no. is $-5+3 i$ and $5-3 i$.